Question: $f(t)=t^2+19t+60$ 1) What are the zeros of the function? Write the smaller $t$ first, and the larger $t$ second. $\text{smaller }t=$
To find the zeros of the function, we need to solve the equation $f(t)=0$. We can do that by factoring $f(t)$. $\begin{aligned} t^2+19t+60&=0 \\\\ (t+15)(t+4)&=0 \\\\ t+15=0&\text{ or }t+4=0 \\\\ t={-15}&\text{ or }t={-4} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $t$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }t\text{-coordinate}&=\dfrac{({-15})+({-4})}{2} \\\\ &={-\dfrac{19}{2}} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $f\left({-\dfrac{19}{2}}\right)$ : $\begin{aligned} f\left({-\dfrac{19}{2}}\right)&=\left({-\dfrac{19}{2}}\right)^2+19\left({-\dfrac{19}{2}}\right)+60 \\\\ &=\dfrac{361}{4}-\dfrac{361}{2}+60 \\\\ &=-\dfrac{121}{4} \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }t&=-15 \\\\ \text{larger }t&=-4 \end{aligned}$ The vertex of the parabola is at $\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$